Question on: JAMB Physics - 2005
a proton moving with a speed of 1.0 x 106ms-1 through a magnetic field of 1.0T experiences a magnetic force of magnitude 8.0 x 10-14 N. The angle between the proton''s velocity and the field is
A
60°
B
90°
C
30°
D
45°
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Correct Option: C
The force acting on moving charge in magnetic field is given by: F=q VBSin θ
sin θ =F=(8.0x10)-14/1.6x10-19x1.0x106x1.0 = (8.0x10)14/16 x 1013
(8.0x 10)-14/16x10-14
=0.5000
∴ θ= sin-1 0.5000
=30°
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